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1

I'm not sure if I understand the question fully, so forgive me if I am incorrect. But doesn't the ball simply bounce up and down, going nowhere in the grand scheme of things, so I would assume that the average normal force is just combatting the force of gravity, making the answer simply "mg".


0

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...


2

The ship and cannonballs will both have the same velocity in the end. The easiest way to think about this from a reference frame in which the ship is stationary before it intersects the cannonballs. The cannonballs will be moving toward the ship in this frame. When the cannonballs reach the springs on the ship their kinetic energy will be converted into ...


0

The cases, assuming the spheres are not rigid: Collision is perfectly elastic=they will only touch Inelastic collision=particles can get displaced because of the compression caused Plastic collision= spheres would stick to each other, the particles will be displaced. Thanks


0

They will not touch. This prevented due to the Pauli exclusion principle. Also, the electrical repulsion of the electron prevents this. Momentum is conserved, however.


1

In an elastic collision involving two objects both momentum and energy are conserved. You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore one object will have the opposite momentum of the other. Since energy is the square of the momentum divided by twice the mass, this means ...


0

Answer to the title of your question : "NO!"... First of all,in general, in most of the mechanics questions, you should make it clear that what is your system and what conservation laws are applicable on it. Like if you have a system on which some constant net external force is acting throughout the time duration you are analyzing the system, you can not ...


2

My question is do the byproducts(elementary particles such as higgs boson) emit photons as they are produced to be filmed on tape by the sensitive detectors? If you're imagining that the particles are emitting photons as the leave the collision center and that the tracks of the particles on the computer screen were recorded by collecting these ...


3

.... We are told that all the subsequent collisions involving the balls and floor are elastic. We are asked to determine the maximum height to which the small sphere will rise on the rebound. The problem does not mention any radii, but if we did know the radius of each of the spheres, would it be valid to bypass conservation of linear ...


4

Here is how you deal this this problem as a system of equations. For each contact pair assign a normal direction $\hat{n}_k$ and and impulse $J_k$. The possible contacts are AB, AC, and AD. We can introduce symmetries and simplifications later. The initial velocity if body A is $v_A$ along the horizontal axis, and after the collision it is $v_A + \Delta ...


5

The problem is equivalent to 4 spheres colliding simultaneously, where top sphere center is at $60^o$ relative to the $x'x$ axes (same goes for bottom sphere): We'll name them: sphere A (dark blue), and spheres 1, 2, and 3. During the collision the spheres will behave like springs with an infinite hook constant. The forces on the spheres will be ...


0

Its pretty much solvable. Lets name the sides of the Hexagon in the order 1,2,3,4. When the hexagon A collides with B,C,D the 2-3 side collide with C and C moves in a direction of A and it will not effect all the other hexagons. Same way the side 1-2 will collide with D and give it motion in a direction perpendicular to side 1-2 and same for side 3-4... ...


0

This is an inelastic collision (the bullet and wood do not elastically bounce off of each other but instead, 'stick' together) and so, kinetic energy is not conserved in the collision. However, the total momentum just before the collision must equal the total momentum just after the collision. $$p(0-) = m_b v_b = p(0+) = (m_b + m_w)v_{bw}$$ Thus, the ...


2

I was thinking about building my virtual elastic bodies as systems of "nodes", each with a certain mass, interconnected by springs. You just described Finite Element Modeling - the cornerstone of mechanical engineering, and the method used for making sure that that bridge won't collapse when an 18 wheeler passes over it. This is an extremely well ...


1

The problem is ambiguous and has infinite solutions. The reason is that now you cannot ignore the relative forces between the A and B,C and D. For instance, if you solve the problem as non simultaneous collisions you will get a different answer depending on the order of the collisions. Suppose that A collides first with C, then A will stop, C will move at ...


-4

This may follow Einstein's equation and may appear to fit into classical picture, but this is taking place between two particles obeying Fermi Dirac statistics. Hence, it is a quantum phenomenon, and the direction of photon emission is arbitrary, as required by the fundamental assumption of quantum mechanics.


20

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...


2

When you compute the final velocity of the parcel you have forgotten that it's no longer traveling at 37 degrees - that was the angle at the end of the chute. While it drops, the horizontal component of velocity doesn't change - it is still $3.4\cdot \cos 37° = 2.71 m/s$. With that, you should be able to solve this.



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